# Integral over scaling function

## Main Question or Discussion Point

Hi,

$\phi(x)$ is an interpolating scaling function (also called fundamental function or Dubuc-Deslauriers function) as given on pages 155 to 158 in these lecture notes: http://pages.unibas.ch/comphys/comphys/TEACH/WS07/course.pdf [Broken]

Why does the follwoing yield:

$\int_{-\infty}^{\infty}\phi(x) dx = 1?$

At least, I assume this yields, because otherwise I cannot show the equality of the first equation of the exercise on page 158 of the above lecture notes:

http://img577.imageshack.us/img577/304/capturevm.png [Broken]

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mathman
φ(x) seems to be the Dirac delta function. Check it out on Wikipedia if you are not familiar with it. Most authors use δ(x).

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Stephen Tashi
Hi,
Why does the follwoing yield:
$\int_{-\infty}^{\infty}\phi(x) dx = 1?$
I don't think that is true.

I only got a hazy grasp of those notes, but, thinking about the case in 1 dimension, the constant function f(x) = 1, when scaled as $f(x)\phi(x)$ is still the constant function 1. (I haven't mastered the notation. By $\phi(x)$, I mean a function such that when x is between the integers k and k + 1, $\phi(x)$ is the sum of the interpolating scaling function centered at k and the interpolating scaling function centered at k + 1.) So one can say that the integral of $(1)( \phi(x))$ from k-1/2 to k+1/2 is 1.

Stephen Tashi
To get my notation straight, I'll use $\phi(x)$ to mean a (single) scaling function centered at 0 which vanishes for x < -1/2 and x > 1/2. (I don't know where the scaling functions in those notes are supposed to vanish. This is just to test an idea.)
If g(x) is a function that is zero except at integers, then I gather the idea of interpolating scaling functions is extend g to a continuous function f(x). Let's say that for x between the integer values k and k + 1, we compute f(x) by adding two functions. The first function $f_{k,1}(x)$ is g(k) times the scaling function $\phi(x-k)$ and the second function $f_{k,2}(x)$ is g(k+1) times the scaling function $\phi(x - (k+1) )$.